Re: de moivre ques involving tan

I never thought I would say this but Plato is wrong. Perhaps he was thinking the cube is on the "cosine" and "sine" separately rather than cubing the entire complex number.

Since this is titled "DeMoivre", , so that . Going back to "rectangular form" that is .

"Imaginary part over real part" is , not .

But to answer the original question, . Multiplying the real part of that by we get and multiplying the imaginary part by we get . That is, the real part of the cube is , the denominator of the fraction on the right.

Multiplying by we get and multiplying by by we get . That is, the imaginary part of the cube is , the numerator of the fraction on the right.

You can also get that result from , using the identities and [tex]cos(a+ b)= cos(a)cos(b)- sin(a)sin(b). Taking , and . Then taking , , , the numerator of the fraction on the right. , the denominator of the fraction on the right.

Re: de moivre ques involving tan

I never thought I would say this but Plato is wrong. Perhaps he was thinking the cube is on the "cosine" and "sine" separately rather than cubing the entire complex number.

Well the title says tangent.
Using a CAS means the cube of the imaginary part. If you notice I did add the extra grouping symbols in that reply.
So as I read it the imaginary part of the numerator is cubed and the real part of the denominator is cubed. That may be wrong but it is the convention I am use to. That does give us the tangent of the title.